A tennis ball is dropped from 1.54 m above the ground. It rebounds to a height of 0.845 m.
With what velocity does it hit the ground? The acceleration of gravity is 9.8 m/s^2 . (Let down be negative.)
Answer in units of m/s.
Well, isn't that quite the bouncy tennis ball! I must say, it clearly has some impressive skills in the rebound game. Now, let's figure out its velocity when it hits the ground, shall we?
We can use the principles of conservation of energy to solve this question. We know that the potential energy at the top is converted into kinetic energy at the bottom, neglecting any loss due to air resistance or friction.
The potential energy at the top (when the ball is dropped) is given by the formula: m * g * h, where m is the mass of the ball, g is the acceleration due to gravity, and h is the height.
The kinetic energy at the bottom (when the ball hits the ground) is given by: (1/2) * m * v^2, where v is the velocity.
Since the potential energy at the top is equal to the kinetic energy at the bottom, we can set up the equation:
m * g * h = (1/2) * m * v^2
Since the mass of the ball cancels out, we're left with:
g * h = (1/2) * v^2
Now, let's plug in the given values. The height, h, is 1.54 m, and the acceleration due to gravity, g, is 9.8 m/s^2.
9.8 m/s^2 * 1.54 m = (1/2) * v^2
Simplifying that equation, we get:
15.092 m^2/s^2 = v^2
To find the velocity, we take the square root of both sides:
v = √(15.092 m^2/s^2)
Calculating that, we find the velocity to be approximately:
v ≈ 3.885 m/s
So, the tennis ball hits the ground with a velocity of approximately 3.885 m/s. I hope my explanation didn't make you bounce off the wall with confusion!
To solve this problem, we can use the principle of conservation of mechanical energy. The total mechanical energy of the system is conserved, assuming there are no external forces acting on the tennis ball.
The mechanical energy of the ball consists of its potential energy when it is at a certain height and its kinetic energy when it is in motion.
Initially, the ball is dropped from a height of 1.54 m. At this point, all of its energy is in the form of potential energy, given by:
Potential Energy = mass × gravity × height
Final velocity is the velocity with which the ball hits the ground. Since the ball rebounds, the final velocity after rebounding is negative due to the change in direction. Therefore, we need to use the negative value for velocity.
Final Kinetic Energy = - Potential Energy
Using the formula for mechanical energy conservation:
Initial Potential Energy = Final Kinetic Energy
mass × gravity × initial height = -(0.5 × mass × velocity^2)
Mass cancels out:
gravity × initial height = -0.5 × velocity^2
Simplifying further:
velocity^2 = -2 × gravity × initial height
Plugging in the given values:
velocity^2 = -2 × 9.8 m/s^2 × 1.54 m
velocity^2 = -30.2888
Taking the square root of both sides to solve for velocity:
velocity = √(-30.2888)
Since velocity is the speed at which the ball hits the ground, the negative value is not meaningful in this context. Therefore, the velocity of the ball as it hits the ground is approximately equal to the square root of 30.2888 m/s, or approximately 5.50 m/s.
To find the velocity with which the tennis ball hits the ground, we can use the principle of conservation of energy.
The initial potential energy of the ball when it is at a height of 1.54 m above the ground can be calculated using the formula:
PE initial = m * g * h
Where:
m = mass of the ball
g = acceleration due to gravity (9.8 m/s^2)
h = height
The final potential energy of the ball when it rebounds to a height of 0.845 m can be calculated using the same formula:
PE final = m * g * h'
Where:
h' = final height
Since the ball rebounds to the same height from which it was dropped, the difference in potential energy between the initial and final positions is zero:
PE final - PE initial = 0
This can be written as:
m * g * h' - m * g * h = 0
Simplifying this equation, we get:
m * g * (h' - h) = 0
Now, we can solve for the velocity of the ball when it hits the ground. The velocity can be obtained from the kinetic energy, which is given by the equation:
KE = (1/2) * m * v^2
Where:
v = velocity
The initial kinetic energy of the ball is zero since it is dropped from rest. The final kinetic energy of the ball just before it hits the ground can be calculated by using the principle of conservation of energy:
KE initial + PE initial = KE final + PE final
Since the ball is at a height of zero above the ground when it hits the ground, the final potential energy is zero. Hence, this equation simplifies to:
PE initial = KE final
Substituting the value of PE initial from the previous equation, we get:
m * g * h = (1/2) * m * v^2
Now, we can solve for v:
v = sqrt(2 * g * h)
Substituting the values of g (9.8 m/s^2) and h (1.54 m) into the equation, we get:
v = sqrt(2 * 9.8 * 1.54)
Solving this equation, we find:
v ≈ 5.71 m/s
Therefore, the velocity with which the tennis ball hits the ground is approximately 5.71 m/s.
V^2 = Vo^2 + 2g*h = 0 + 19.6*1.54 = 30.2
V = 5.49 m/s.